A question on The Nature of Gravity in relation to close proximity to earth

Newton's law of universal gravitation states that every mass in the universe attracts every other mass with a force proportional to the product of their masses.

See website for diagrams: http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html

Inverse Square Law, Gravity

As one of the fields which obey the general inverse square law, the gravity field can be put in the form shown below, showing that the acceleration of gravity, g, is an expression of the intensity of the gravity field.
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Unquote:

Place the aircraft above the surface of the Earth to the elavation of 10000ft. The distance from the Earth will calculate how much of the Earth's sphere it is exposed to and therefore the measure of gravitational pull it is feeling from the Earth. This measurement will not be inversed as the craft is now the point of focus. It will be the opposite of inversed. Now measure the aircraft's mass weight. It's weight inversed by the measurement of distance will calculate the amount of gravitational pull it is effecting on the Earth. Take the measurement of gravitational pull felt by the craft from the Earth and the measurement of gravitational pull felt by the Earth from the craft. Add these together. This is the increase in gravitational pull that an aircraft of this weight will experience at this elevation above this planet. (This obviously does not take into acount lift, induced drag, thrust, pressure, temperature or density. Only weight, elevation and gravity drag. )

If this is not the case, can anyone explain why please?
 

K^2

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Doesn't matter how much of Earth is exposed. Gravity cannot be shielded. You are always attracted by the pull of the entire Earth's mass. It doesn't matter that most of it is not exposed to you.
 

DaveC426913

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...The distance from the Earth will calculate how much of the Earth's sphere it is exposed to...
This statement is nonsensical. Any object is "exposed to" the entirety of the Earth at all times. Gravity acts universally, whether there is something in the way or not.

Every single atom of the Earth - from the near side through the far side - pulls on the airplane with a force that is directly proportional to the atom's mass and inversely proportional to the square of its distance. Full stop.
 

russ_watters

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Also note that that equation is for really for point masses (and, iirc, also works for uniform spherical masses). Since earth is not a point mass and doesn't have a uniform mass distribution, the accuracy of that equation actually increases as you move away from the earth.
 
This statement is nonsensical. Any object is "exposed to" the entirety of the Earth at all times. Gravity acts universally, whether there is something in the way or not.

Every single atom of the Earth - from the near side through the far side - pulls on the airplane with a force that is directly proportional to the atom's mass and inversely proportional to the square of its distance. Full stop.
Thanks for your answer Dave.
The airplane's atoms also pull on the Earth with a force that is directly proportional to the atom's mass and inversely proportional to the square of it's distance.
Thinking of myself now as a "hover person" if I hover 1foot above the earth, will I be using the same force of thrust at 1foot elevation as I will at 2000ft?
 
Also note that that equation is for really for point masses (and, iirc, also works for uniform spherical masses). Since earth is not a point mass and doesn't have a uniform mass distribution, the accuracy of that equation actually increases as you move away from the earth.
Thank you for your reply Russ.
Please excuse me, I am a leyman in physics terms. Your reply interests me. Could you possibly elaborate on the differences between "point mass" and "uniform mass" and what you mean by the accuracy of that equation actually increasing as you move away from the Earth?
 
Doesn't matter how much of Earth is exposed. Gravity cannot be shielded. You are always attracted by the pull of the entire Earth's mass. It doesn't matter that most of it is not exposed to you.
Cheers for your reply as well. Duly noted.
 
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Thank you for your reply Russ.
Please excuse me, I am a leyman in physics terms. Your reply interests me. Could you possibly elaborate on the differences between "point mass" and "uniform mass" and what you mean by the accuracy of that equation actually increasing as you move away from the Earth?
A point mass is what it says, it means you can consider the object as a point. A uniform mass distribution, would have an equal density at any point in the object, for the Earth this is not the case, for example differences in the masses of rocks and so on.
This means that as you get further away from the earth, the uneven distribution of mass becomes more negligible, as the error would get smaller as a percentage of the total mass, and hence the force would be closer to the actual force.
 

DaveC426913

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...if I hover 1foot above the earth, will I be using the same force of thrust at 1foot elevation as I will at 2000ft?
No. The gravitational pull of the Earth is a fraction smaller at 20,000 feet. But it s a very small fraction - about 1/10th of a percent. So, instead of (about) 9.8, it would be 9.79.
 
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A point mass is what it says, it means you can consider the object as a point. A uniform mass distribution, would have an equal density at any point in the object, for the Earth this is not the case, for example differences in the masses of rocks and so on.
This means that as you get further away from the earth, the uneven distribution of mass becomes more negligible, as the error would get smaller as a percentage of the total mass, and hence the force would be closer to the actual force.
Thanks for that. Understood and duly noted.
 
No. The gravitational pull of the Earth is a fraction smaller at 20,000 feet. But it s a very small fraction - about 1/10th of a percent. So, instead of (about) 9.8, it would be 9.79.
A commercial jet plane will fly faster with increased fuel economy at optimum cruising altitude:
Despite the fact that due to lack of oxygen (low air density or high density altitude) the engine is not performing at optimum power.
Despite the fact that the minimal decrease in gravity at this level of altitude will have little effect on a plane.
Despite the fact that low air density (or high density altitude) will cause a lower rate of buoyancy.
Despite the fact that low air density (or high density altitude) will cause the wings to produce less lift.
Despite the fact that whilst thrust specific fuel comsumption is high, actual thrust out put is decreased.

Can this just be due to less drag? If so, are we talking about induced drag or gravity drag?
A jet plane in take off will, at the same level of thrust, travel more slowly at a steeper angle of ascent.
Its thrust specific fuel consumption will be far higher, at lower speeds, in higher air density (or low density altitude), during it's ascent. Although jet engines are designed to be more efficient in-taking extremely cold air found at altitude, the oxygen rich, high density air at lower altitude will afford optimum fuel combustion.
If a jet engine has high thrust specific fuel consumption at altitude due to an air to fuel mix ratio, why does a jet plane have a high thrust specific fuel consumption at lower speed during an ascent?
Can this just be attributed to increased drag due to air density?
 

DaveC426913

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A commercial jet plane will fly faster with increased fuel economy at optimum cruising altitude...

Can this just be due to less drag? If so, are we talking about induced drag or gravity drag?
Yes, the biggest factor in efficiency of jet travel is drag. They go to higher altitudes because of the reduced drag from thinner air.

I do not know what you mean by 'gravity drag'.
 

D H

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Can this just be due to less drag? If so, are we talking about induced drag or gravity drag?
Answered in order, yes and no. Parasitic drag, not induced drag or gravity drag, is the driving factor here. As parasitic drag is proportional to density, it is better to fly high where density is low.
 

D H

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I do not know what you mean by 'gravity drag'.
It's a term used in rocketry, not avionics. The concept applies to landing on or taking off from any planetary body, including the Moon. It has nothing to do with atmosphere.

Here are a couple of ways to get a satellite from the surface of the Earth to GEO:
  1. Install the vehicle as a payload on a rocket. The rocket, usually in two stages, launches into low earth orbit. From LEO, a third stage performs a burn to put the vehicle in an orbit whose apogee is at GEO altitude. The vehicle drifts until it comes close to apogee, at which time it performs another burn to insert the vehicle into geosynchronous orbit.
  2. Just thrust straight up from the desired subsatellite point on equator, gradually picking up horizontal speed to keep the desired subsatellite point directly below. No need for that high-G, high velocity rocketry stuff, right? The maximum speed needed with this approach is less than half that needed for a low Earth orbit.
The problem with the latter approach is that the amount of fuel required is huge. This approach has a very high gravity drag term.
 
Yes, the biggest factor in efficiency of jet travel is drag. They go to higher altitudes because of the reduced drag from thinner air.
DH edit. This was a mess. For one thing, looked like DaveC posted a lot of info on Cesnas; he did not. The link was completely wrong, for another. I think this is what the poster meant to say.


(A Personal Opinion)
The change of gravity is negligible. The actual mass of the aircraft however (and thus its weight) is considerably reduced throughout the flight as its fuel is burnt. This is much more important than any change in gravity due to altitude.

As for the relation between engine performance and altitude, here are typical numbers in the case of a propeller-driven aircraft (taken out of a 1978 Model Cessna 172 owner's handbook), running at constant 2200 RPM, 15degC - 2degC per 1000ft (in this case at least, the colder air does not compensate for the lower density):

100 knots @ 2000ft
99 knots @ 4000ft
98 knots @ 6000ft
97 knots @ 8000ft
96 knots @ 10000ft
95 knots @ 12000ft

As for the relation between climb performance (feet per minute) and altitude:

675FPM @ 2000ft
580FPM @ 4000ft
485FPM @ 6000ft
390FPM @ 8000ft
295FPM @ 10000ft
200FPM @ 12000ft

Note that the relations are typical for a specific single engine prop plane and has perhaps nothing to do with a jet engine.
Additional information Cessna 172:
Cruising Speed = 122 knots
Service Ceiling = 14200ft
Although a Cessna 172 is most certainly not a jet plane, there are similarity's in what happens for both during an ascent and again during cruising altitude. Although almost maximum power is used both at altitude cruising speed and in take off, take off is achieved at lower speeds than ascent or cruising speeds. The initial ascent happens quite fast, then ascent slows, the speed slows, and the ascent slows again as does the speed. Then the aircraft reaches cruising altitude and it all levels out. The altitude this happens at depends on how heavy the aircraft is and tweaking is possible dependent on thrust versus weight versus altitude versus engine power.
Why is the aircraft experiencing such an escalation in drag between 2000ft and 12000ft?
Gravity tails off very gradually and would have little effect on an aircraft except for the fact that the aircraft is also a mass. Newton's law of universal gravitation states that every mass in the universe attracts every other mass with a force proportional to the product of their masses.


I do not know what you mean by 'gravity drag'.
Quote: Wikapedia
In astrodynamics and rocketry, gravity drag (or gravity losses) is a measure of the loss in the net performance of a rocket while it is thrusting in a gravitational field.

DH edit: Rest of wikipedia quote deleted.
 
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[*]Just thrust straight up from the desired subsatellite point on equator, gradually picking up horizontal speed to keep the desired subsatellite point directly below. No need for that high-G, high velocity rocketry stuff, right? The maximum speed needed with this approach is less than half that needed for a low Earth orbit.[/list]
The problem with the latter approach is that the amount of fuel required is huge. This approach has a very high gravity drag term.
Very interesting. Do you have or know where to get any data charting gravity losses in relation to velocity? I am particularly interested in the region of 0 to 35000ft.
Does anyone know where to find known density altitude parameters for temperature and air pressure charts?
 
Answered in order, yes and no. Parasitic drag, not induced drag or gravity drag, is the driving factor here. As parasitic drag is proportional to density, it is better to fly high where density is low.
I am not disputing this fact. However during my trawling around pilot forums I noticed a distinct lack of understanding as to this phenomenon due to the factors that I outlined as "despite". That a lack of air density, whilst inducing less parasitic drag also causes there to be less lift, less bouyancy, less efficient fuel consumption per thrust output, yet the craft flies faster with more fuel efficiency. Or could there be another factor behind the high fuel consumption during ascent. I suppose I'll have to get the dreaded maths involved, but in the mean time...
 
Apologies to Editor: Didn't know it was going to come out like that. Still learning your page parameters. Noted: Will keep quotes more to the point.
 

D H

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Using gravity drag here is essentially wrong in this context, Time Machine. You already are taking lift and weight into account. Adding in gravity drag is double booking lift versus weight.
 

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